Maximum Picture Viewing Angle
by: Joshua Wood
Problem: A 4 by 4 picture hangs on a wall such that its bottom edge is 2 ft above your eye
level. How far back from the picture should you stand, directly in front of the picture, in
order to view the picture under the maximum angle?
Solution: Let the distance x and angles a and b be as in the figure below.
Thus tan(a + b) =
6
x
⇒ a + b = tan−1 x6 . Also we have tan b =
a = tan−1
2
x
6
6
2
− b = tan−1 − tan−1
x
x
x
⇒ b = tan−1 x2 . So
Therefore
da
1
−6
1
−2
=
−
36 ·
4 ·
2
dx
1 + x2 x
1 + x2 x2
−6
2
= 2
+ 2
x + 36 x + 4
Setting da/dx = 0 gives us
6
2
= 2
+ 36
x +4
3
1
= 2
2
x + 36
x +4
x2
3x2 + 12 = x2 + 36
2x2 = 24
x2 = 12
√
x = 2 3
Now we just need to establish that this value of x gives us a maximum angle a.
a = a(x) is a continuously differentiable function on (0, ∞). Note that
lim a =
x→∞
lim (tan−1
x→∞
6
2
− tan−1 )
x
x
= tan−1 0 − tan−1 0
= 0
=
π π
−
2
2
=
lim+ (tan−1
=
x→0
lim a
x→0+
2
2
6
− tan−1 )
x
x
Also, by the monotonicity of arctangent we have that for all x ∈ (0, ∞), a(x) > 0, since
6
2
6
2
> ⇒ tan−1 > tan−1
x
x
x
x
(this is apparent from the physical setup of the problem as well). Thus we have a positive,
continuously differentiable function on (0, ∞) that limits to zero on both ends with one
√
critical value. Therefore, the critical value x = 2 3 ft, gives us a maximum viewing angle.
3